Hệ đã cho vô nghiệm khi:

\(\dfrac{1}{2}=\dfrac{m}{3m-1}\ne\dfrac{6}{3}\)

\(\Rightarrow3m-1=2m\)

\(\Rightarrow m=1\)

\(\left\{{}\begin{matrix}2x+y=3m-1\\x-2y=-m-3\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\\dfrac{3m-1-y}{2}-2y=-m-3\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\3m-1-y-4y=-2m-6\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\5y=5m+5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\y=m+1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-m-1}{2}\\y=m+1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m-1\\y=m+1\end{matrix}\right.\)

Vậy hpt trên có nghiệm duy nhất \(\left\{{}\begin{matrix}x=m-1\\y=m+1\end{matrix}\right.\)

Ta có: y = x² \(\Leftrightarrow\) m + 1 = (m - 1)² \(\Leftrightarrow\) m + 1 = m² - 2m + 1

\(\Leftrightarrow\) m² - 3m = 0

\(\Leftrightarrow\) m(m - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=0\\m-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=0\\m=3\end{matrix}\right.\)

Vậy m = 0; m = 3 thì hpt trên có nghiệm duy nhất và thỏa mãn y = x²

Chúc bn học tốt!

\(1;\left\{{}\begin{matrix}mx+2y=7\\2x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7-mx}{2}\\2x+\dfrac{3\left(7-mx\right)}{2}=5\left(1\right)\end{matrix}\right.\) 

\(hệ\) \(pt\) \(có\) \(nghiệm\) \(duy\) \(nhất\Leftrightarrow\left(1\right)có\) \(ngo\) \(duy\) \(nhất\)

\(\left(1\right)\Leftrightarrow\dfrac{4x+3\left(7-mx\right)}{2}=5\Leftrightarrow4x+21-3mx=10\Leftrightarrow x\left(4-3m\right)=-11\)

\(với:m\ne\dfrac{4}{3}\) \(thì\) \(hpt\) \(có\) \(ngo\) \(duy-nhất\left(x;y\right)=\left\{\dfrac{-11}{4-3m};\dfrac{7-m\left(\dfrac{-11}{4-3m}\right)}{2}\right\}\)

\(2,\left\{{}\begin{matrix}2x-y=m\\-4x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x-m\\-4x+2\left(2x-m\right)=4\left(1\right)\end{matrix}\right.\)

hệ pt vô nghiệm khi (1) vô nghiệm

(1)\(\Leftrightarrow-4x+4x-2m=4\Leftrightarrow m=-2\Rightarrow m=-2\)

thì hệ pt có vô số nghiệm

\(\Rightarrow m\ne-2\) thì hpt vô nghiệm

Để phương trình có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)

=>\(m-1\ne2m\)

=>\(m\ne-1\)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-2xm+m^2+5m=3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(m-1-2m\right)=-m^2-5m+3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(-m-1\right)=-m^2-2m-1=-\left(m+1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\cdot\left(-1\right)\cdot\left(m+1\right)=-\left(m+1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\end{matrix}\right.\)

\(x^2-y^2=24\)

=>\(\left(m+1\right)^2-\left(m-3\right)^2=24\)

=>\(m^2+2m+1-m^2+6m-9=24\)

=>8m-8=24

=>m=4(nhận)

Để hệ có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)

=>\(2m\ne m-1\)

=>\(m\ne-1\)(1)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(m-1\right)-2mx+m^2+5m-3m+1=0\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(-m-1\right)+m^2+2m+1=0\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(m+1\right)=\left(m+1\right)^2\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\end{matrix}\right.\)

\(x^2-y^2< 4\)

=>\(\left(m+1\right)^2-\left(m-3\right)^2< 4\)

=>\(m^2+2m+1-m^2+6m-9< 4\)

=>8m-8<4

=>8m<12

=>\(m< \dfrac{3}{2}\)

Kết hợp (1), ta được: \(\left\{{}\begin{matrix}m< \dfrac{3}{2}\\m\ne-1\end{matrix}\right.\)